Risk of ruin for HUSNG's - Gambling and Probability

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Risk of ruin for HUSNG's

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06-13-2018 , 11:59 AM

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Hi guys,

Could anyone help me calculate my risk of ruin for HUSNG's at different bankrolls with a 57% WR assuming I played an infinite number of games and the bankroll grows to a point where going broke is all but impossible?

Would be interesting to see the % for say 1 buy in, 10, 20, 50, 100 etc.

Is there a formula I can use?

Many thanks.

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06-13-2018 , 02:56 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

I'm pretty sure I can do this but first:

What is the buy in (prize pool amt.) and fee?

What does the 57% represent? I assume it's the percent of times you win, but maybe not. If it is the winning frequency I can calculate your ROI and variance and answer the question. If it is ROI, I can also do the problem.

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06-13-2018 , 04:00 PM

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Sorry forgot to mention rake, 3% rake so for a $100 game it's $97+3. 57% is win rate, think it equals about a 10.5% ROI.

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06-13-2018 , 06:04 PM

statmanhal

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Join Date: Jan 2009 Posts: 4,987

The risk of ruin approximate formula is

RoR = e^(-2BW/V)

where

B = bankroll
W = win rate $ (a function of the win probability of 57% and prize structure)
V = variance of win rate $ [(std dev)^2]

For a 97 + 3 total buy in (3% fee, wow) and a 57% win probability your ROI is 10.58%. Your standard deviation calculates to 120.34.

Here are the results for 100 tournaments and 1000 tournaments

Bankroll.	RoR-100	RoR-1000
1	84.8%	86.4%
5	42.5%	48.2%
10	16.4%	23.2%
25	0.5%	2.6%
50	0%	0.1%
75	0%	0%
100	0%	0%

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06-13-2018 , 07:32 PM

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Cool, thanks a lot! Surprised that 1 buy in risk of ruin is under 90%.

What does e stand for in your formula?

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06-13-2018 , 08:37 PM

RustyBrooks

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Join Date: Feb 2006 Posts: 24,647

"e" is a mathematical constant, it's the base of natural logarithms

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06-13-2018 , 08:44 PM

statmanhal

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Join Date: Jan 2009 Posts: 4,987

e ~ 2.718

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06-13-2018 , 08:49 PM

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Edit: ignore me.

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06-13-2018 , 08:55 PM

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Ok I'm really confused, sorry. Where does the number of games come into things with this? I tried calculating 2.718^(-2(1)(57)/120.34^2) for one buy in bankroll and the answer was 2.71.

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06-13-2018 , 09:01 PM

#10

RustyBrooks

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Join Date: Feb 2006 Posts: 24,647

You need to have your win rate and bankroll in the same units I think. You have BR=1 but your win rate is in dollars (i.e. you have 57 in that variable). You either need to say BR=$100 or WR=.57

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06-13-2018 , 09:04 PM

#11

RustyBrooks

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Join Date: Feb 2006 Posts: 24,647

Although also I think 57 is the wrong number for your win rate - that should be win rate in $/game which since you have a 10.5% ROI I guess would be $10.5?

ETA: your variance also has to be in the same units as your WR and BR

ETA 2: I realize I should probably specify what I mean
W = 10.58
S = 120.34
(V = S*S)
B = 100

With those numbers I get 86.40 as your RoR. That formula is considering that you will "play forever" so it's like the number of tournaments is infinite. I don't know what method was used to arrive at the numbers for 100 and 1000 games but there's probably a modified formula for that.

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06-13-2018 , 09:49 PM

#12

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

Quote:

Originally Posted by RustyBrooks

. That formula is considering that you will "play forever" so it's like the number of tournaments is infinite. I don't know what method was used to arrive at the numbers for 100 and 1000 games but there's probably a modified formula for that.

Correct. I believe I used the formula from Don Blankenship's Blackjack Attack book which he called the short term risk of ruin formula. I could probably verify that but my result was gotten from a program I wrote some 9 years ago and I'm too lazy to do that.

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06-13-2018 , 10:33 PM

#13

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Quote:

Originally Posted by RustyBrooks

Please can you write out the formula using the figures? When I try using the ones you've used, 2.718^(-2(100)(10.58)/14481.71), I get an answer of 2.3 or something. Thanks

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06-13-2018 , 11:09 PM

#14

RustyBrooks

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Join Date: Feb 2006 Posts: 24,647

e^(-2*100*10.58/14481.7156)
e^(-2116.0/14481.7156)
e^-0.146115284849
2.718^-0.146115284849
0.864071175092

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06-13-2018 , 11:13 PM

#15

RustyBrooks

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The only thing I can think of is that for some reason you're doing
e^(1-0.1461) = 2.3485

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06-14-2018 , 01:53 AM

#16

DianeAbbott

adept

Join Date: Mar 2018 Posts: 860

Well I really appreciate both of your time to help me with this this, thanks.

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